Date: 12/08/98 at 13:39:59 From: ACHILLE Subject: Binomial theorem and expansion. 1) Expand (1+x+x^2)^n, n a positive integer, in ascending powers of x as far as the term in x^3. The answer is: 1 + nx + (n(n+1)x^2)/2 + (n(n-1)(n+4)x^3)/6 I can't obtain the given answer from this formula. Is there another one? Formula: (1+x)^n = 1 + nx + (n(n-1)/(1*2))x^2 + ... + (n(n-1)...(n-r+1))/(r(r-1) ... 3*2*1)x^r + ... x^n if n belongs to R. How can I proceed? 2) In the expansion of (1+x+px^2)^7, the coefficient of x^2 is zero. Find the value of p. The given answer is -3. 3) Given that (1+ax+bx^2)^10 = 1 - 30x + 410x^2 ..., find the values of the constants a, b. The given answer is a = -3, b = 1/2. I tried to proceed by taking 10(ax + bx^2) = -30x and obtained a correctly but not b. Thanks, merci beaucoup! Achilles K.
Date: 12/08/98 at 16:24:36 From: Doctor Ezra Subject: Re: Binomial theorem and expansion. Dear Achilles, Thank you for writing Dr. Math. What you need is not the binomial theorem, which expands expressions of the form (x+y)^n, but the *trinomial* theorem, which expands expressions of the form (x+y+z)^n. That theorem is as follows: If we write C(n;p,q,r) = n!/(p!q!r!), then: (x+y+z)^n = sum C(n;p,q,r) x^p * y^q * z^r where the sum is taken over all ordered triples (a, b, c) such that a + b + c = n. Take the expression (1 + ax + bx^2)^5 (that's not what you asked, but it'll do for an example.) Use the trinomial theorem with n = 5, x = 1, y = ax and z=bx^2, and you'll add up all the expressions: C(5;p,q,r) * 1^p * (ax)^q * (bx^2)^r It turns out that there are 21 terms in this expression. One term, for example, is: C(5;2,1,2) * 1^2 * (ax)^1 * (bx^2)^2 = 30 * ax * b^2x^4 = 30abx^5 Hope this helps. Bonne chance! - Doctor Ezra, The Math Forum http://mathforum.org/dr.math/
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